| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Easy -1.2 This is a straightforward projectile motion question requiring only standard formulas and basic calculations. Part (i) uses Pythagoras, part (ii) applies standard SUVAT equations for flight time and range, and part (iii) involves simple verification and a conceptual observation about complementary angles. All techniques are routine for M1 students with no problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Speed \(= \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\) m s\(^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical: \(0 = 20t - \frac{1}{2}(10)t^2\) | M1 | Using \(s=ut+\frac{1}{2}at^2\) or \(v=u+at\) for flight time |
| \(t(20 - 5t) = 0\), so \(t = 4\) s | A1 | |
| \(R = 15 \times 4 = 60\) m | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Flight time: \(0 = 15t - 5t^2 \Rightarrow t = 3\) s | M1 | |
| \(R = 20 \times 3 = 60\) m. Same range. | A1 | Shown convincingly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No, not always the same | B1 | |
| The range is the same only when the two components are swapped (complementary angles). For arbitrary directions with speed 25 m s\(^{-1}\) the range varies. | B1 | Must give justification, e.g. range formula \(R = \frac{u^2 \sin 2\theta}{g}\) depends on angle |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Speed $= \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25$ m s$^{-1}$ | B1 | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical: $0 = 20t - \frac{1}{2}(10)t^2$ | M1 | Using $s=ut+\frac{1}{2}at^2$ or $v=u+at$ for flight time |
| $t(20 - 5t) = 0$, so $t = 4$ s | A1 | |
| $R = 15 \times 4 = 60$ m | M1 A1 | |
## Part (iii)(A):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Flight time: $0 = 15t - 5t^2 \Rightarrow t = 3$ s | M1 | |
| $R = 20 \times 3 = 60$ m. Same range. | A1 | Shown convincingly |
## Part (iii)(B):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No, not always the same | B1 | |
| The range is the same only when the two components are swapped (complementary angles). For arbitrary directions with speed 25 m s$^{-1}$ the range varies. | B1 | Must give justification, e.g. range formula $R = \frac{u^2 \sin 2\theta}{g}$ depends on angle |
---2 In this question, air resistance should be neglected.\\
Fig. 2 illustrates the flight of a golf ball. The golf ball is initially on the ground, which is horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-2_273_1109_1297_479}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
It is hit and given an initial velocity with components of $15 \mathrm {~ms} ^ { - 1 }$ in the horizontal direction and $20 \mathrm {~ms} ^ { - 1 }$ in the vertical direction.
\begin{enumerate}[label=(\roman*)]
\item Find its initial speed.
\item Find the ball's flight time and range, $R \mathrm {~m}$.
\item (A) Show that the range is the same if the components of the initial velocity of the ball are $20 \mathrm {~ms} ^ { - 1 }$ in the horizontal direction and $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the vertical direction.\\
(B) State, justifying your answer, whether the range is the same whenever the ball is hit with the same initial speed.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 2013 Q2 [8]}}